The value of machine depreciates
every year at the rate of 10% p.a. If
the present value of machine is Rs
6561. What was its value 3 year ago?
Answers
Answer:
Rate = - 10% p.a. ( The rate is negative since the price is depriciating.)
Time = 3 years
Final price = Rs. 6561
To Find :
Initial price
Solution :
Let initial price = x
Now, We know that,
\underline{\boxed{\sf Final \: Price = Initial \: Price \Bigg( 1 + \dfrac{Rate}{100} \Bigg) ^{time}}}
FinalPrice=InitialPrice(1+
100
Rate
)
time
By putting values,
: \implies \sf Rs. \: 6561 = x \times \Bigg( 1 + \dfrac{- 10}{100} \Bigg) ^{3}:⟹Rs.6561=x×(1+
100
−10
)
3
: \implies \sf Rs. \: 6561 = x \times \Bigg( 1 - \dfrac{ 1\cancel{0}}{10\cancel{0}} \Bigg) ^{3}:⟹Rs.6561=x×(1−
10
0
1
0
)
3
: \implies \sf Rs. \: 6561 = x \times \Bigg( \dfrac{10}{10} - \dfrac{ 1}{10} \Bigg) ^{3}:⟹Rs.6561=x×(
10
10
−
10
1
)
3
: \implies \sf Rs. \: 6561 = x \times \Bigg( \dfrac{10 - 1}{10} \Bigg) ^{3}:⟹Rs.6561=x×(
10
10−1
)
3
: \implies \sf Rs. \: 6561 = x \times \Bigg( \dfrac{9}{10} \Bigg) ^{3}:⟹Rs.6561=x×(
10
9
)
3
: \implies \sf Rs. \: 6561 = x \times \dfrac{9}{10} \times \dfrac{9}{10} \times \dfrac{9}{10}:⟹Rs.6561=x×
10
9
×
10
9
×
10
9
: \implies \sf Rs. \: \cancel{6561}^{729} \times \dfrac{10}{\cancel{9}} \times \dfrac{10}{9} \times \dfrac{10}{9} = x:⟹Rs.
6561
729
×
9
10
×
9
10
×
9
10
=x
: \implies \sf Rs. \: \cancel{729}^{81} \times 10 \times \dfrac{10}{\cancel{9}} \times \dfrac{10}{9} = x:⟹Rs.
729
81
×10×
9
10
×
9
10
=x
: \implies \sf Rs. \: \cancel{81}^{9} \times 10 \times 10 \times \dfrac{10}{\cancel{9}} = x:⟹Rs.
81
9
×10×10×
9
10
=x
: \implies \sf Rs. \: 9 \times 10 \times 10 \times 10= x:⟹Rs.9×10×10×10=x
: \implies \sf Rs. \: 9000= x:⟹Rs.9000=x
: \implies \sf x = Rs. \: 9000:⟹x=Rs.9000
\large \underline{\boxed{\sf x = Rs. \: 9000}}
x=Rs.9000
So, the price of the machine 3 years ago = Rs. 9000.
Step-by-step explanation:
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